An element 'X' (At mass = `40 "g mol"^(-1)`) having fcc structure, has unit cell length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g in 'X' `(N_A=6.022xx10^23 "mol"^(-1))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Density of Element 'X'

1. **Identify the parameters:**
- Molar mass (m) of element 'X' = 40 g/mol
- Unit cell length (a) = 400 pm = \(400 \times 10^{-10}\) cm
- For FCC structure, the number of atoms per unit cell (Z) = 4

2. **Calculate the volume of the unit cell (V):**
\[ V = a^3 = (400 \times 10^{-10} \text{ cm})^3 = 6.4 \times 10^{-23} \text{ cm}^3 \]

3. **Use the density formula:**
\[ \text{Density} (\rho) = \frac{Z \cdot m}{N_A \cdot V} \]
where \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\).

4. **Substituting the values:**
\[ \rho = \frac{4 \cdot 40 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1} \cdot 6.4 \times 10^{-23} \text{ cm}^3} \]

5. **Calculating the density:**
\[ \rho = \frac{160 \text{ g}}{6.022 \times 10^{23} \cdot 6.4 \times 10^{-23}} = \frac{160}{3.84528} \approx 4.15 \text{ g/cm}^3 \]

### Step 2: Calculate the Number of Unit Cells in 4 g of 'X'

1. **Calculate the number of moles in 4 g of 'X':**
\[ \text{Number of moles} = \frac{\text{mass}}{m} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol} \]

2. **Calculate the number of unit cells:**
\[ \text{Number of unit cells} = \text{Number of moles} \times N_A \times Z \]
\[ = 0.1 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} \times 4 \]

3. **Calculating the number of unit cells:**
\[ = 0.1 \times 6.022 \times 10^{23} \times 4 = 2.4088 \times 10^{23} \approx 2.41 \times 10^{22} \]

### Final Answers:
- **Density of 'X'**: \(4.15 \text{ g/cm}^3\)
- **Number of unit cells in 4 g of 'X'**: \(2.41 \times 10^{22}\)

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